Answer
$B = 1.32\times 10^{-7}~T$
Work Step by Step
We can find the magnetic field at $P_2$:
$dB = \frac{\mu_0}{4\pi}~\frac{i~ds~sin~\theta}{r^2}$
$B = \int_{0}^{L}~\frac{\mu_0}{4\pi}~\frac{i~ds~sin~\theta}{r^2}$
$B = \int_{0}^{L}~\frac{\mu_0}{4\pi}~\frac{i~ds~R^2}{(s^2+R^2)^{3/2}}$
$B = \frac{\mu_0~i}{4\pi~R}~\frac{s}{\sqrt{s^2+R^2}}~\Big \vert_{0}^{L}$
$B = \frac{\mu_0~i}{4\pi~R}~(\frac{L}{\sqrt{L^2+R^2}}-0)$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.693~A)}{(4\pi)~(0.251~m)}~\frac{0.136~m}{\sqrt{(0.136~m)^2+(0.251~m)^2}}$
$B = 1.32\times 10^{-7}~T$
